The first term of an arithmetic progression is 1 and the sum of the first nine terms equal to 369. The first and the ninth term of a geometic progression colncide with the first and the ninth term of the arithmetic progression. Find the seventh term of the geometric progression.

Answer

To solve the problem step by step, we will follow the given information about the arithmetic progression (AP) and the geometric progression (GP).

### Step 1: Understand the given information
- The first term of the arithmetic progression (AP) is \( a = 1 \).
- The sum of the first 9 terms of the AP is \( S_9 = 369 \).

### Step 2: Use the formula for the sum of the first n terms of an AP
The formula for the sum of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
where \( n \) is the number of terms, \( a \) is the first term, and \( d \) is the common difference.

### Step 3: Substitute the known values into the formula
For our case, \( n = 9 \), \( a = 1 \), and \( S_9 = 369 \):
\[ 369 = \frac{9}{2} \times (2 \cdot 1 + (9-1)d) \]
This simplifies to:
\[ 369 = \frac{9}{2} \times (2 + 8d) \]

### Step 4: Solve for \( d \)
Multiply both sides by 2 to eliminate the fraction:
\[ 738 = 9 \times (2 + 8d) \]
Now divide both sides by 9:
\[ 82 = 2 + 8d \]
Subtract 2 from both sides:
\[ 80 = 8d \]
Now divide by 8:
\[ d = 10 \]

### Step 5: Find the ninth term of the AP
The \( n \)-th term of an AP is given by:
\[ a_n = a + (n-1)d \]
For the ninth term (\( n = 9 \)):
\[ a_9 = 1 + (9-1) \cdot 10 = 1 + 80 = 81 \]

### Step 6: Relate the GP to the AP
The first term of the GP is \( b_1 = 1 \) (same as the first term of the AP) and the ninth term of the GP is \( b_9 = 81 \). The formula for the \( n \)-th term of a GP is:
\[ b_n = b_1 \cdot r^{n-1} \]
For the ninth term:
\[ b_9 = b_1 \cdot r^8 \]
Substituting the known values:
\[ 81 = 1 \cdot r^8 \]
Thus:
\[ r^8 = 81 \]

### Step 7: Solve for the common ratio \( r \)
Taking the eighth root of both sides:
\[ r = 81^{1/8} \]
Since \( 81 = 3^4 \):
\[ r = (3^4)^{1/8} = 3^{4/8} = 3^{1/2} = \sqrt{3} \]

### Step 8: Find the seventh term of the GP
Using the formula for the \( n \)-th term of the GP:
\[ b_7 = b_1 \cdot r^{6} \]
Substituting the known values:
\[ b_7 = 1 \cdot (\sqrt{3})^6 = (\sqrt{3})^6 = 3^{3} = 27 \]

### Final Answer
The seventh term of the geometric progression is \( \boxed{27} \).

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